description/proof of that for map between finite-product-of-copies-of-field vectors spaces, map is linear iff map is represented by matrix w.r.t. canonical bases
Topics
About: vectors space
The table of contents of this article
Starting Context
- The reader knows a definition of %field name% vectors space.
- The reader knows a definition of product vectors space.
- The reader knows a definition of linear map.
- The reader knows a definition of %ring name% matrices space.
Target Context
- The reader will have a description and a proof of the proposition that for any map between any finite-product-of-copies-of-field vectors spaces, the map is linear if and only if the map is represented by the matrix with respect to the canonical bases.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(F\): \(\in \{\text{ the fields }\}\)
\(d_1\): \(\in \mathbb{N} \setminus \{0\}\)
\(d_2\): \(\in \mathbb{N} \setminus \{0\}\)
\(F^{d_1}\): \(= \text{ the } d_1 \text{ -dimensional } F \text{ vectors space }\)
\(F^{d_2}\): \(= \text{ the } d_2 \text{ -dimensional } F \text{ vectors space }\)
\(B_1\): \(= \{b_{1, 1}, ..., b_{1, d_1}\}\), \(= \text{ the canonical basis for } F^{d_1}\)
\(B_2\): \(= \{b_{2, 1}, ..., b_{2, d_2}\}\), \(= \text{ the canonical basis for } F^{d_2}\)
\(f\): \(: F^{d_1} \to F^{d_2}\)
\(M\): \(= \begin{pmatrix} f (b_{1, 1})^1 & ... & f (b_{1, d_1})^1 \\ ... \\ f (b_{1, 1})^{d_2} & ... & f (b_{1, d_1})^{d_2} \end{pmatrix}\)
//
Statements:
\(f \in \{\text{ the linear maps }\}\)
\(\iff\)
\(\forall v \in F^{d_1} (f (v) = M v^t)\)
//
"the canonical basis" means that \(b_{l, j} = (0, ..., 0, 1, 0, ..., 0)^t\) where \(1\) is the \(j\)-th component.
2: Note
Someone may wonder whether this is only with respect to the canonical bases, but of course not so, but the existence of the representative matrix for arbitrary bases is derived based on this canonical case: refer to the definition of representative matrix of linear map between finite-dimensional vectors spaces with respect to bases.
3: Proof
Whole Strategy: Step 1: suppose that \(f\) is linear; Step 2: see that \(\forall v \in F^{d_1} (f (v) = M v^t)\); Step 3: suppose that \(\forall v \in F^{d_1} (f (v) = M v^t)\); Step 4: see that \(f\) is linear.
Step 1:
Let us suppose that \(f\) is linear.
Step 2:
Let \(v \in F^{d_1}\) be any.
\(v = v^j b_{1, j}\).
\(f (v) = f (v^j b_{1, j}) = v^j f (b_{1, j})\), because \(f\) is linear, \(= v^j f (b_{1, j})^l b_{2, l} = \begin{pmatrix} v^j f (b_{1, j})^1 \\ ... \\ v^j f (b_{1, j})^{d_2} \end{pmatrix} = \begin{pmatrix} f (b_{1, 1})^1 & ... & f (b_{1, d_1})^1 \\ ... \\ f (b_{1, 1})^{d_2} & ... & f (b_{1, d_1})^{d_2} \end{pmatrix} \begin{pmatrix} v^1 \\ ... \\ v^{d_1} \end{pmatrix} = M v^t\).
Step 3:
Let us suppose that \(\forall v \in F^{d_1} (f (v) = M v^t)\).
Let \(v, v' \in \mathbb{R}^{d_1}\) and \(r, r' \in \mathbb{R}\) be any.
\(f (r v + r' v') = M (r v + r' v')^t = M (r v^t + r' v'^t) = r M v^t + r' M v'^t\), which we accept as a well-known fact, \(= r f (v) + r' f (v')\).
So, \(f\) is linear.
